Continuous-time Fourier transform and properties

Why it matters

Most measured signals are not periodic — a transient from an impact test, a single pulse from a proximity sensor, a decaying response. The continuous-time Fourier transform (CTFT) gives their spectrum, which tells you how much bandwidth an amplifier or ADC must have, how a filter will change the signal, and how a delay or modulation shows up in frequency.

Key ideas

From series to transform. Let the period of a periodic signal grow to infinity: the harmonic spacing ω0 shrinks to zero, the line spectrum becomes continuous, and the Fourier series becomes the Fourier transform. X(jω) is a density: it has the unit of x multiplied by seconds (e.g. V·s, or V/Hz).

Two conventions. In angular frequency ω (rad/s) the inverse transform has a 1/2π factor. In cyclic frequency f (Hz) there is no 2π factor in either direction. Questions use both; always check which one is in use.

Existence. Dirichlet conditions (absolute integrability plus finitely many discontinuities and extrema) are sufficient. All energy signals have a transform. Power signals such as 1, cos ω0t, u(t) and periodic signals have transforms only if impulses are allowed.

Spectrum of a real signal. If x(t) is real, X(−jω) = X*(jω): the magnitude spectrum is even and the phase spectrum is odd. A real and even signal has a real and even transform; a real and odd signal has an imaginary and odd transform.

Time–frequency duality and the uncertainty idea. Narrow in time means wide in frequency: a pulse of width τ has its main lobe out to 1/τ Hz. This is why fast edges need wide-band instruments.

Key properties (x(t) ↔ X(jω))

• Linearity.
• Time shift: x(t − t0) ↔ X(jω) e^(−jωt0) — magnitude unchanged, linear phase added.
• Frequency shift (modulation): x(t) e^(jω0t) ↔ X(j(ω − ω0)); x(t) cos ω0t ↔ ½[X(j(ω − ω0)) + X(j(ω + ω0))].
• Time scaling: x(at) ↔ (1/|a|) X(jω/a).
• Time reversal: x(−t) ↔ X(−jω).
• Differentiation: dx/dt ↔ jω X(jω); integration: ∫ from −∞ to t of x ↔ X(jω)/(jω) + π X(0) δ(ω).
• Convolution: x * h ↔ X(jω) H(jω); multiplication: x(t)·y(t) ↔ (1/2π) X * Y.
• Duality: if x(t) ↔ X(jω) then X(jt) ↔ 2π x(−ω).
• Area: X(0) = ∫ x(t) dt and x(0) = (1/2π) ∫ X(jω) dω.

Frequency response. For an LTI system with impulse response h(t), H(jω) is the CTFT of h(t) (the Laplace transform on the jω-axis for a stable system). A sinusoid cos ω0t comes out as |H(jω0)| cos(ω0t + ∠H(jω0)).

Formulas

• X(jω) = ∫ x(t) e^(−jωt) dt — analysis equation (ω form).
• x(t) = (1/2π) ∫ X(jω) e^(jωt) dω — synthesis equation.
• X(f) = ∫ x(t) e^(−j2πft) dt, x(t) = ∫ X(f) e^(j2πft) df — f form.
• E = ∫ |x(t)|² dt = (1/2π) ∫ |X(jω)|² dω = ∫ |X(f)|² df — Parseval (Rayleigh) energy theorem.
• e^(−at)u(t) ↔ 1/(a + jω), a > 0.
• e^(−a|t|) ↔ 2a/(a² + ω²), a > 0.
• A·rect(t/τ) ↔ Aτ·sinc(fτ) with sinc(u) = sin(πu)/(πu); in ω form Aτ·sin(ωτ/2)/(ωτ/2).
• δ(t) ↔ 1, 1 ↔ 2π δ(ω) (= δ(f)), cos ω0t ↔ π[δ(ω − ω0) + δ(ω + ω0)], u(t) ↔ π δ(ω) + 1/(jω).

Symbols: ω angular frequency (rad/s); f frequency (Hz); t0 delay (s); a decay rate (s⁻¹); τ pulse width (s); A amplitude (unit of x); X spectral density (unit of x·s); E normalised energy.

Worked examples

Example 1 (standard). For x(t) = e^(−2t) u(t), find |X(jω)| and ∠X(jω) at ω = 2 rad/s, and the fraction of the signal energy in the band |ω| ≤ 2 rad/s.

1. X(jω) = 1/(2 + jω).
2. At ω = 2: |X| = 1/√(2² + 2²) = 1/√8 = 0.354; ∠X = −tan⁻¹(2/2) = −45°.
3. Total energy (time domain): E = ∫ from 0 to ∞ of e^(−4t) dt = 1/4.
4. Band energy by Parseval: E_band = (1/2π) ∫ from −2 to 2 of dω/(4 + ω²) = (1/2π) · 2 · (1/2) tan⁻¹(1) = (1/2π)(π/4) = 1/8.
5. Fraction = (1/8)/(1/4). |X(j2)| ≈ 0.354, phase −45°, and 50 % of the energy lies within |ω| ≤ 2 rad/s (ω = a is the half-power frequency).

Example 2 (GATE level). A sensor emits a 2 V rectangular pulse 1 ms wide. (a) Find X(0), the first spectral null and the energy. (b) Use Parseval to evaluate ∫ [sin(πt)/(πt)]² dt over all t.

1. (a) X(f) = Aτ sinc(fτ), A = 2 V, τ = 1 × 10⁻³ s.
2. X(0) = Aτ = 2 × 10⁻³ V·s (the area of the pulse).
3. First null where fτ = 1: f = 1/τ = 1000 Hz.
4. Energy E = A²τ = 4 × 10⁻³ V²·s. Delaying the pulse changes only the phase, not any of these.
5. (b) sinc(t) = sin(πt)/(πt) is the transform pair of rect(f) (height 1, width 1 Hz) by duality.
6. Parseval: ∫ sinc²(t) dt = ∫ |rect(f)|² df = 1 × 1. (a) X(0) = 2 mV·s, first null 1 kHz, E = 4 × 10⁻³ V²·s; (b) the integral equals 1.

Common mistakes

• Mixing the ω and f conventions: 1 ↔ 2πδ(ω) but 1 ↔ δ(f).
• Forgetting the 1/|a| amplitude factor in time scaling.
• Thinking a time delay changes |X|. It only adds phase −ωt0.
• Dropping the πX(0)δ(ω) term when integrating or when transforming u(t).
• Writing sinc(x) without stating whether it means sin(πx)/(πx) or sin(x)/x.
• Claiming e^(−at)u(t) with a < 0 has a Fourier transform — it grows, so the integral diverges.

For GATE IN

Expect property-based MCQs (time shift, scaling, modulation, duality), NAT questions that use Parseval or X(0) = ∫x dt to evaluate an integral without integrating, the transform of a pulse or exponential with a magnitude or phase at a given frequency, and the steady-state output of an LTI system for a sinusoidal input. Practise recognising the standard pairs and using area and energy shortcuts.

Quick check

1. What is the CTFT of δ(t − 3)?
2. If x(t) ↔ X(jω), what is the transform of x(3t)?
3. What is ∫ x(t) dt in terms of the spectrum?
4. Is the spectrum of a real, even signal real, imaginary or complex?
5. The first null of a 0.2 ms rectangular pulse is at what frequency? Answers: 1. e^(−j3ω). 2. (1/3) X(jω/3). 3. X(0). 4. Real (and even). 5. 5 kHz.